Stability of the Ishikawa iteration scheme with errors for two strictly hemicontractive operators in Banach spaces

Abstract

The main purpose of this paper is to establish the convergence, almost common-stability and common-stability of the Ishikawa iteration scheme with error terms in the sense of Xu (J. Math. Anal. Appl. 224:91-101, 1998) for two Lipschitz strictly hemicontractive operators in arbitrary Banach spaces.

1 Preliminaries

Let K be a nonempty subset of an arbitrary Banach space E and $E^{\ast }$ be its dual space. The symbols $D(T)$, $R(T)$ and $F(T)$ stand for the domain, the range and the set of fixed points of T respectively (for a single-valued map $T:X\to X$, $x\in X$ is called a fixed point of T iff $T(x)=x$). We denote by J the normalized duality mapping from E to $2^{E^{\ast }}$ defined by

$$J(x)=\{f^{\ast }\in E^{\ast }:〈x,f^{\ast }〉={\parallel x\parallel }^2={\parallel f^{\ast }\parallel }^2\}.$$

Let T be a self-mapping of K.

Definition 1 Then T is called Lipshitzian if there exists $L>0$ such that

$$\parallel Tx-Ty\parallel ⩽L\parallel x-y\parallel$$

(1.1)

for all $x,y\in K$. If $L=1$, then T is called non-expansive, and if $0⩽L<1$, T is called contraction.

Definition 2[2, 3]

• 1. The mapping T is said to be pseudocontractive if the inequality

  $$\parallel x-y\parallel ⩽\parallel x-y+t((I-T)x-(I-T)y)\parallel$$

  (1.2)

holds for each $x,y\in K$ and for all $t>0$. As a consequence of a result of Kato [4], it follows from the inequality (1.2) that T is pseudocontractive if and only if there exists $j(x-y)\in J(x-y)$ such that

$$〈Tx-Ty,j(x-y)〉⩽{\parallel x-y\parallel }^2$$

(1.3)

for all $x,y\in K$.

• 2. T is said to be strongly pseudocontractive if there exists a $t>1$ such that

  $$\parallel x-y\parallel \le \parallel (1+r)(x-y)-rt(Tx-Ty)\parallel$$

  (1.4)

for all $x,y\in D(T)$ and $r>0$.

• 3. T is said to be local strongly pseudocontractive if, for each $x\in D(T)$, there exists a $t_x>1$ such that

  $$\parallel x-y\parallel \le \parallel (1+r)(x-y)-r{t}_x(Tx-Ty)\parallel$$

  (1.5)

for all $y\in D(T)$ and $r>0$.

• 4. T is said to be strictly hemicontractive if $F(T)\ne \phi$ and if there exists a $t>1$ such that

  $$\parallel x-q\parallel \le \parallel (1+r)(x-q)-rt(Tx-q)\parallel$$

  (1.6)

for all $x\in D(T)$, $q\in F(T)$ and $r>0$.

It is easy to verify that an iteration scheme ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ which is T-stable on K is almost T-stable on K. Osilike [5] proved that an iteration scheme which is almost T-stable on X may fail to be T-stable on X.

Clearly, each strongly pseudocontractive operator is local strongly pseudocontractive.

Chidume [6] established that the Mann iteration sequence converges strongly to the unique fixed point of T in case T is a Lipschitz strongly pseudo-contractive mapping from a bounded closed convex subset of $L_p$ (or $l_p$) into itself. Afterwards, several authors generalized this result of Chidume in various directions. Chidume [7] proved a similar result by removing the restriction ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$. Tan and Xu [8] extended that result of Chidume to the Ishikawa iteration scheme in a p-uniformly smooth Banach space. Chidume and Osilike [2] improved the result of Chidume [6] to strictly hemicontractive mappings defined on a real uniformly smooth Banach space.

Recently, some researchers have generalized the results to real smooth Banach spaces, real uniformly smooth Banach spaces, real Banach spaces; or to the Mann iteration method, the Ishikawa iteration method; or to strongly pseudocontractive operators, local strongly pseudocontractive operators, strictly hemicontractive operators [9–19].

The main purpose of this paper is to establish the convergence, almost common-stability and common-stability of the Ishikawa iteration scheme with error terms in the sense of Xu [1] for two Lipschitz strictly hemicontractive operators in arbitrary Banach spaces. Our results extend, improve and unify the corresponding results in [2, 3, 10, 11, 15–18, 20–25].

2 Main results

We need the following results.

Lemma 3[26]

Let${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{{\gamma }_n\}}_{n=0}^{\mathrm{\infty }}$and${\{{\omega }_n\}}_{n=0}^{\mathrm{\infty }}$be nonnegative real sequences such that

$${\alpha }_{n+1}\le (1-{\omega }_n){\alpha }_n+{\omega }_n{\beta }_n+{\gamma }_n,n\ge 0,$$

with${\{{\omega }_n\}}_{n=0}^{\mathrm{\infty }}\subset [0,1]$, ${\sum }_{n=0}^{\mathrm{\infty }}{\omega }_n=\mathrm{\infty }$, ${\sum }_{n=0}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$and${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$. Then${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$.

Lemma 4[27]

Let${\{a_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=0}^{\mathrm{\infty }}$be sequences of nonnegative real numbers and$0\le \theta <1$, so that

$$a_{n+1}\le \theta a_n+b_n,\mathit{\text{for all}}n\ge 0.$$

• (i) If ${lim}_{n\to \mathrm{\infty }}{b}_n=0$, then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

• (ii) If ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$, then ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n<\mathrm{\infty }$.

Lemma 5[4]

Let$x,y\in X$. Then$\parallel x\parallel \le \parallel x+ry\parallel$for every$r>0$if and only if there is$f\in J(x)$such that$Re〈y,f〉\ge 0$.

Lemma 6[2]

Let$T:D(T)\subseteq X\to X$be an operator with$F(T)\ne \mathrm{\varnothing }$. Then T is strictly hemicontractive if and only if there exists$t>1$such that for all$x\in D(T)$and$q\in F(T)$, there exists$j\in J(x-q)$satisfying

$$Re〈x-Tx,j(x-q)〉\ge (1-\frac{1}{t}){\parallel x-q\parallel }^2.$$

Lemma 7[24]

Let X be an arbitrary normed linear space and$T:D(T)\subseteq X\to X$be an operator.

• (i) If T is a local strongly pseudocontractive operator and $F(T)\ne \mathrm{\varnothing }$, then $F(T)$ is a singleton and T is strictly hemicontractive.

• (ii) If T is strictly hemicontractive, then $F(T)$ is a singleton.

In the sequel, let $k=\frac{t-1}{t}\in (0,1)$, where t is the constant appearing in (1.6). Further L denotes the common Lipschitz constant of T and S, and I denotes the identity mapping on an arbitrary Banach space X.

Definition 8 Let K be a nonempty convex subset of X and $T,S:K\to K$ be two operators. Assume that $x_o\in K$ and $x_{n+1}=f(T,S,x_n)$ defines an iteration scheme which produces a sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}\subset K$. Suppose, furthermore, that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to $q\in F(T)\cap F(S)\ne \phi$. Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any bounded sequence in K and put ${\epsilon }_n=\parallel y_{n+1}-f(T,S,y_n)\parallel$.

• (i) The iteration scheme ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ defined by $x_{n+1}=f(T,S,x_n)$ is said to be common-stable on K if ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$.

• (ii) The iteration scheme ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ defined by $x_{n+1}=f(T,S,x_n)$ is said to be almost common-stable on K if ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$.

We now establish our main results.

Theorem 9 Let K be a nonempty closed convex subset of an arbitrary Banach space X and$T,S:K\to K$be two Lipschitz strictly hemicontractive operators. Suppose that${\{u_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$are arbitrary bounded sequences in K, and${\{a_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{c_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{a_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$and${\{c_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$are any sequences in$[0,1]$satisfying

• (i) $a_n+b_n+c_n=1=a_n^{\mathrm{\prime }}+b_n^{\mathrm{\prime }}+c_n^{\mathrm{\prime }}$,

• (ii) $c_n^{\mathrm{\prime }}=o(b_n^{\mathrm{\prime }})$,

• (iii) ${lim}_{n\to \mathrm{\infty }}{c}_n=0$,

• (iv) ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n^{\mathrm{\prime }}=\mathrm{\infty }$,

• (v) $L[{(1+L)}^2{b}_n^{\mathrm{\prime }}+c_n^{\mathrm{\prime }}+(1+L)(b_n+c_n)]+\frac{c_n^{\mathrm{\prime }}}{b_n^{\mathrm{\prime }}}\le k(k-s)$, $n\ge 0$,

where s is a constant in$(0,k)$. Suppose that${\{x_n\}}_{n=0}^{\mathrm{\infty }}$is the sequence generated from an arbitrary$x_0\in K$by

$$\begin{array}{r}{x}_{n+1}=a_n^{\mathrm{\prime }}{x}_n+b_n^{\mathrm{\prime }}T{z}_n+c_n^{\mathrm{\prime }}{v}_n,\\ z_n=a_n{x}_n+b_{n}S{x}_n+c_n{u}_n,n\ge 0.\end{array}$$

(2.1)

Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any sequence in K and define ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ by

$${\epsilon }_n=\parallel y_{n+1}-p_n\parallel ,n\ge 0,$$

where

$$\begin{array}{r}{p}_n=a_n^{\mathrm{\prime }}{y}_n+b_n^{\mathrm{\prime }}T{w}_n+c_n^{\mathrm{\prime }}{v}_n,\\ w_n=a_n{y}_n+b_{n}S{y}_n+c_n{u}_n,n\ge 0.\end{array}$$

(2.2)

Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to the common fixed point q of T and S. Also,

  $$\begin{array}{rcl}\parallel x_{n+1}-q\parallel & \le & (1-s{b}_n^{\mathrm{\prime }})\parallel x_n-q\parallel \\ +L(1+L)k^{-1}{b}_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel +(1+L)k^{-1}{c}_n^{\mathrm{\prime }}\parallel v_n-q\parallel ,n\ge 0,\end{array}$$

  (b)

  $$\begin{array}{rcl}\parallel y_{n+1}-q\parallel & \le & (1-s{b}_n^{\mathrm{\prime }})\parallel y_n-q\parallel \\ +L(1+L)k^{-1}{b}_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel +(1+L)k^{-1}{c}_n^{\mathrm{\prime }}\parallel v_n-q\parallel +{\epsilon }_n,n\ge 0,\end{array}$$

• (c) ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$, so that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is almost common-stable on K,

• (d) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

Proof From (ii), we have $c_n^{\mathrm{\prime }}=t_n{b}_n^{\mathrm{\prime }}$, where $t_n\to 0$ as $n\to \mathrm{\infty }$. It follows from Lemma 7 that $F(T)\cap F(S)$ is a singleton; that is, $F(T)\cap F(S)=\{q\}$ for some $q\in K$. Set

$$M=max\{\underset{n\ge 0}{sup}\{\parallel u_n-q\parallel \},\underset{n\ge 0}{sup}\{\parallel v_n-q\parallel \}\}.$$

Since T is strictly hemicontractive, it follows form Lemma 6 that

$$Re〈x-Tx,j(x-q)〉\ge k{\parallel x-q\parallel }^2,\mathrm{\forall }x\in K,$$

which implies that

$$Re〈(I-T-kI)x-(I-T-kI)q,j(x-q)〉\ge 0,\mathrm{\forall }x\in K.$$

In view of Lemma 5, we have

$$\parallel x-q\parallel \le \parallel x-q+r[(I-T-kI)x-(I-T-kI)q]\parallel ,\mathrm{\forall }x\in K,\mathrm{\forall }r>0.$$

(2.3)

Also,

$$\begin{array}{rcl}(1-b_n^{\mathrm{\prime }})x_n& =& (1-(1-k)b_n^{\mathrm{\prime }})x_{n+1}+b_n^{\mathrm{\prime }}(I-T-kI)x_{n+1}\\ +b_n^{\mathrm{\prime }}(T{x}_{n+1}-T{z}_n)-c_n^{\mathrm{\prime }}(v_n-x_n),\end{array}$$

(2.4)

and

$$(1-b_n^{\mathrm{\prime }})q=(1-(1-k)b_n^{\mathrm{\prime }})q+b_n^{\mathrm{\prime }}(I-T-kI)q.$$

(2.5)

From (2.4) and (2.5), we infer that for all $n\ge 0$,

$$\begin{array}{rcl}(1-b_n^{\mathrm{\prime }})\parallel x_n-q\parallel & \ge & \parallel (1-(1-k)b_n^{\mathrm{\prime }})(x_{n+1}-q)+b_n^{\mathrm{\prime }}(I-T-kI)(x_{n+1}-q)\parallel \\ -b_n^{\mathrm{\prime }}\parallel T{x}_{n+1}-T{z}_n\parallel -c_n^{\mathrm{\prime }}\parallel v_n-x_n\parallel \\ =& (1-(1-k)b_n^{\mathrm{\prime }})\parallel x_{n+1}-q+\frac{b_n^{\mathrm{\prime }}}{1-(1-k)b_n^{\mathrm{\prime }}}(I-T-kI)(x_{n+1}-q)\parallel \\ -b_n^{\mathrm{\prime }}\parallel T{x}_{n+1}-T{z}_n\parallel -c_n^{\mathrm{\prime }}\parallel v_n-x_n\parallel \\ \ge & (1-(1-k)b_n^{\mathrm{\prime }})\parallel x_{n+1}-q\parallel -b_n^{\mathrm{\prime }}\parallel T{x}_{n+1}-T{z}_n\parallel \\ -c_n^{\mathrm{\prime }}\parallel v_n-x_n\parallel ,\end{array}$$

which implies that for all $n\ge 0$,

(2.6)

(2.7)

(2.8)

Substituting (2.8) in (2.7), we have

$$\begin{array}{rcl}\parallel x_{n+1}-z_n\parallel & \le & [b_n^{\mathrm{\prime }}+c_n^{\mathrm{\prime }}+(1+L)b_n+c_n]\parallel x_n-q\parallel \\ +L{b}_n^{\mathrm{\prime }}[[1+(1+L)b_n+c_n]\parallel x_n-q\parallel \\ +c_n\parallel u_n-q\parallel ]+c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +c_n\parallel u_n-q\parallel \\ =& [(1+L)b_n^{\mathrm{\prime }}+L(1+L)b_n{b}_n^{\mathrm{\prime }}+(1+L)b_n+c_n^{\mathrm{\prime }}\\ +(1+L{b}_n^{\mathrm{\prime }})c_n]\parallel x_n-q\parallel \\ +c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +(1+L{b}_n^{\mathrm{\prime }})c_n\parallel u_n-q\parallel .\end{array}$$

(2.9)

Substituting (2.9) in (2.6), we get

$$\begin{array}{rcl}\parallel x_{n+1}-q\parallel & \le & (1-k{b}_n^{\mathrm{\prime }}+k^{-1}{c}_n^{\mathrm{\prime }})\parallel x_n-q\parallel +k^{-1}L{b}_n^{\mathrm{\prime }}\left[\right[(1+L)b_n^{\mathrm{\prime }}\\ +L(1+L)b_n{b}_n^{\mathrm{\prime }}+(1+L)b_n+c_n^{\mathrm{\prime }}+(1+L{b}_n^{\mathrm{\prime }})c_n]\parallel x_n-q\parallel \\ +c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +(1+L{b}_n^{\mathrm{\prime }})c_n\parallel u_n-q\parallel ]+k^{-1}{c}_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ =& [1-b_n^{\mathrm{\prime }}[k-k^{-1}L((1+L)b_n^{\mathrm{\prime }}+L(1+L)b_n{b}_n^{\mathrm{\prime }}\\ +(1+L)b_n+c_n^{\mathrm{\prime }}+(1+L{b}_n^{\mathrm{\prime }})c_n)-k^{-1}{t}_n]]\parallel x_n-q\parallel \\ +k^{-1}L{b}_n^{\mathrm{\prime }}(1+L{b}_n^{\mathrm{\prime }})c_n\parallel u_n-q\parallel +k^{-1}(1+L{b}_n^{\mathrm{\prime }})c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ \le & [1-b_n^{\mathrm{\prime }}[k-k^{-1}L({(1+L)}^2{b}_n^{\mathrm{\prime }}+(1+L)b_n\\ +c_n^{\mathrm{\prime }}+(1+L)c_n)-k^{-1}{t}_n]]\parallel x_n-q\parallel \\ +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ \le & (1-s{b}_n^{\mathrm{\prime }})\parallel x_n-q\parallel +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ \le & (1-s{b}_n^{\mathrm{\prime }})\parallel x_n-q\parallel +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_{n}M+k^{-1}(1+L)b_n^{\mathrm{\prime }}{t}_{n}M\\ =& (1-s{b}_n^{\mathrm{\prime }})\parallel x_n-q\parallel +k^{-1}(1+L)M{b}_n^{\mathrm{\prime }}(L{c}_n+t_n).\end{array}$$

Put

we have

$${\alpha }_{n+1}\le (1-{\omega }_n){\alpha }_n+{\omega }_n{\beta }_n+{\gamma }_n,n\ge 0.$$

Observe that ${\sum }_{n=0}^{\mathrm{\infty }}{\omega }_n=\mathrm{\infty }$, ${\omega }_n\in [0,1]$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$. It follows from Lemma 3 that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$.

We also have

$$\begin{array}{rcl}(1-b_n^{\mathrm{\prime }})y_n& =& (1-(1-k)b_n^{\mathrm{\prime }})p_n+b_n^{\mathrm{\prime }}(I-T-kI)p_n\\ +b_n^{\mathrm{\prime }}(T{p}_n-T{w}_n)-c_n^{\mathrm{\prime }}(v_n-y_n).\end{array}$$

(2.10)

From (2.5) and (2.10), it follows that for all $n\ge 0$,

$$\begin{array}{rcl}(1-b_n^{\mathrm{\prime }})\parallel y_n-q\parallel & \ge & \parallel (1-(1-k)b_n^{\mathrm{\prime }})(p_n-q)+b_n^{\mathrm{\prime }}(I-T-kI)(p_n-q)\parallel \\ -b_n^{\mathrm{\prime }}\parallel T{p}_n-T{w}_n\parallel -c_n^{\mathrm{\prime }}\parallel v_n-y_n\parallel \\ =& (1-(1-k)b_n^{\mathrm{\prime }})\parallel p_n-q\\ +\frac{b_n^{\mathrm{\prime }}}{1-(1-k)b_n^{\mathrm{\prime }}}(I-T-kI)(p_n-q)\parallel \\ -b_n^{\mathrm{\prime }}\parallel T{p}_n-T{w}_n\parallel -c_n^{\mathrm{\prime }}\parallel v_n-y_n\parallel \\ \ge & (1-(1-k)b_n^{\mathrm{\prime }})\parallel p_n-q\parallel -b_n^{\mathrm{\prime }}\parallel T{p}_n-T{w}_n\parallel \\ -c_n^{\mathrm{\prime }}\parallel v_n-y_n\parallel ,\end{array}$$

which implies that for all $n\ge 0$,

(2.11)

(2.12)

(2.13)

Substituting (2.13) in (2.12), we have

$$\begin{array}{rcl}\parallel p_n-w_n\parallel & \le & [b_n^{\mathrm{\prime }}+c_n^{\mathrm{\prime }}+(1+L)b_n+c_n]\parallel y_n-q\parallel \\ +L{b}_n^{\mathrm{\prime }}[[1+(1+L)b_n+c_n]\parallel y_n-q\parallel \\ +c_n\parallel u_n-q\parallel ]+c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +c_n\parallel u_n-q\parallel \\ =& [(1+L)b_n^{\mathrm{\prime }}+L(1+L)b_n{b}_n^{\mathrm{\prime }}+(1+L)b_n+c_n^{\mathrm{\prime }}\\ +(1+L{b}_n^{\mathrm{\prime }})c_n]\parallel y_n-q\parallel \\ +c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +(1+L{b}_n^{\mathrm{\prime }})c_n\parallel u_n-q\parallel .\end{array}$$

(2.14)

Substituting (2.14) in (2.11), we get

$$\begin{array}{rcl}\parallel p_n-q\parallel & \le & (1-k{b}_n^{\mathrm{\prime }}+k^{-1}{c}_n^{\mathrm{\prime }})\parallel y_n-q\parallel +k^{-1}L{b}_n^{\mathrm{\prime }}\left[\right[(1+L)b_n^{\mathrm{\prime }}\\ +L(1+L)b_n{b}_n^{\mathrm{\prime }}+(1+L)b_n+c_n^{\mathrm{\prime }}+(1+L{b}_n^{\mathrm{\prime }})c_n]\parallel y_n-q\parallel \\ +c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +(1+L{b}_n^{\mathrm{\prime }})c_n\parallel u_n-q\parallel ]+k^{-1}{c}_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ =& [1-b_n^{\mathrm{\prime }}[k-k^{-1}L((1+L)b_n^{\mathrm{\prime }}+L(1+L)b_n{b}_n^{\mathrm{\prime }}\\ +(1+L)b_n+c_n^{\mathrm{\prime }}+(1+L{b}_n^{\mathrm{\prime }})c_n)-k^{-1}{t}_n]]\parallel y_n-q\parallel \\ +k^{-1}L{b}_n^{\mathrm{\prime }}(1+L{b}_n^{\mathrm{\prime }})c_n\parallel u_n-q\parallel +k^{-1}(1+L{b}_n^{\mathrm{\prime }})c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ \le & [1-b_n^{\mathrm{\prime }}[k-k^{-1}L({(1+L)}^2{b}_n^{\mathrm{\prime }}+(1+L)b_n\\ +c_n^{\mathrm{\prime }}+(1+L)c_n)-k^{-1}{t}_n]]\parallel y_n-q\parallel \\ +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ \le & (1-s{b}_n^{\mathrm{\prime }})\parallel y_n-q\parallel +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \end{array}$$

(2.15)

for any $n\ge 0$. Thus (2.15) implies that

$$\begin{array}{rcl}\parallel y_{n+1}-q\parallel & \le & \parallel y_{n+1}-p_n\parallel +\parallel p_n-q\parallel \\ \le & (1-s{b}_n^{\mathrm{\prime }})\parallel y_n-q\parallel +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +{\epsilon }_n\\ =& (1-{\omega }_n)\parallel y_n-q\parallel +{\omega }_n{\beta }_n+{\gamma }_n.\end{array}$$

(2.16)

With

we have

$${\alpha }_{n+1}\le (1-{\omega }_n){\alpha }_n+{\omega }_n{\beta }_n+{\gamma }_n,n\ge 0.$$

Observe that ${\sum }_{n=0}^{\mathrm{\infty }}{\omega }_n=\mathrm{\infty }$, ${\omega }_n\in [0,1]$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$. It follows from Lemma 3 that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-q\parallel =0$.

Suppose that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$. It follows from equation (2.15) that

$$\begin{array}{rcl}{\epsilon }_n& \le & \parallel y_{n+1}-q\parallel +\parallel p_n-q\parallel \\ \le & (1-s{b}_n^{\mathrm{\prime }})\parallel y_n-q\parallel +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +\parallel y_{n+1}-q\parallel \to 0,\end{array}$$

as $n\to \mathrm{\infty }$; that is, ${\epsilon }_n\to 0$ as $n\to \mathrm{\infty }$. □

Using the techniques in the proof of Theorem 9, we have the following results.

Theorem 10 Let X, K, T, S, s, ${\{u_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{z_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{w_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$and${\{p_n\}}_{n=0}^{\mathrm{\infty }}$be as in Theorem  9. Suppose that${\{a_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{c_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{a_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$and${\{c_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$are sequences in$[0,1]$satisfying conditions (i), (iii)-(v) of Theorem  9 with

$$\sum _{n=0}^{\mathrm{\infty }}{c}_n^{\mathrm{\prime }}<\mathrm{\infty }.$$

Then the conclusions of Theorem  9 hold.

Theorem 11 Let X, K, T, S, s, ${\{u_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{v_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{z_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{w_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$and${\{p_n\}}_{n=0}^{\mathrm{\infty }}$be as in Theorem  9. Suppose that${\{a_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{c_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{a_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$and${\{c_n^{\mathrm{\prime }}\}}_{n=0}^{\mathrm{\infty }}$are sequences in$[0,1]$satisfying condition (i), (iii) and (v) of Theorem  9 with

where m is a constant. Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to the common fixed point q of T and S. Also,

  $$\parallel x_{n+1}-q\parallel \le (1-sm)\parallel x_n-q\parallel +C,\mathrm{\forall }n\ge 0,$$

where

$$C=k^{-1}(1+L)[\underset{n\ge 0}{Lsup}\{c_n\parallel u_n-q\parallel \}+\underset{n\ge 0}{sup}\{c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \}],$$

(b)

$$\begin{array}{rcl}\parallel y_{n+1}-q\parallel & \le & (1-sm)\parallel y_n-q\parallel +k^{-1}L(1+L)c_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +{\epsilon }_n,\mathrm{\forall }n\ge 0,\end{array}$$

• (c) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

Proof As in the proof of Theorem 9, we conclude that $F(T)\cap F(S)=\{q\}$ and

$$\begin{array}{rcl}\parallel x_{n+1}-q\parallel & \le & (1-s{b}_n^{\mathrm{\prime }})\parallel x_n-q\parallel +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ \le & (1-sm)\parallel x_n-q\parallel +k^{-1}L(1+L)c_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel \\ \le & (1-sm)\parallel x_n-q\parallel +C,\mathrm{\forall }n\ge 0.\end{array}$$

Let

Observe that $0\le \theta <1$ and ${lim}_{n\to \mathrm{\infty }}{b}_n=0$. It follows from Lemma 4 that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$.

Also, from (2.15), we have

$$\begin{array}{rcl}\parallel y_{n+1}-q\parallel & \le & (1-s{b}_n^{\mathrm{\prime }})\parallel y_n-q\parallel +k^{-1}L(1+L)b_n^{\mathrm{\prime }}{c}_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +{\epsilon }_n\\ \le & (1-sm)\parallel y_n-q\parallel +k^{-1}L(1+L)c_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +{\epsilon }_n.\end{array}$$

Suppose that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$. It follows from equation (2.15) that

$$\begin{array}{rcl}{\epsilon }_n& \le & \parallel y_{n+1}-q\parallel +\parallel p_n-q\parallel \\ \le & (1-sm)\parallel y_n-q\parallel +k^{-1}L(1+L)c_n\parallel u_n-q\parallel \\ +k^{-1}(1+L)c_n^{\mathrm{\prime }}\parallel v_n-q\parallel +\parallel y_{n+1}-q\parallel \to 0,\end{array}$$

as $n\to \mathrm{\infty }$; that is, ${\epsilon }_n\to 0$ as $n\to \mathrm{\infty }$.

Conversely, suppose that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$. Put

Observe that $0\le \theta <1$ and ${lim}_{n\to \mathrm{\infty }}{b}_n=0$. It follows from Lemma 4 that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-q\parallel =0$. □

As an immediate consequence of Theorems 9 and 11, we have the following:

Corollary 12 Let K be a nonempty closed convex subset of an arbitrary Banach space X and$T,S:K\to K$be two Lipschitz strictly hemicontractive operators. Suppose that${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=0}^{\mathrm{\infty }}$are any sequences in$[0,1]$satisfying

• (vi) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (vii) $L[{(1+L)}^2{\alpha }_n+(1+L){\beta }_n]\le k(k-s)$, $n\ge 0$,

where s is a constant in$(0,k)$. Suppose that${\{x_n\}}_{n=0}^{\mathrm{\infty }}$is the sequence generated from an arbitrary$x_0\in K$by

Let ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be any sequence in K and define ${\{{\epsilon }_n\}}_{n=0}^{\mathrm{\infty }}$ by

$${\epsilon }_n=\parallel y_{n+1}-p_n\parallel ,n\ge 0,$$

where

$$p_n=(1-{\alpha }_n)y_n+{\alpha }_{n}T{w}_n,$$

and

$$w_n=(1-{\beta }_n)y_n+{\beta }_{n}S{y}_n,n\ge 0.$$

Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to the common fixed point q of T and S,

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_n<\mathrm{\infty }$ implies that ${lim}_{n\to \mathrm{\infty }}{y}_n=q$, so that ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is almost common-stable on K,

• (c) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

Corollary 13 Let X, K, T, S, s, ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{z_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{w_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$and${\{p_n\}}_{n=0}^{\mathrm{\infty }}$be as in Theorem  9. Suppose that${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=0}^{\mathrm{\infty }}$are sequences in$[0,1]$satisfying conditions (vi)-(vii) and (iii) of Theorem  9 with

$${\alpha }_n\ge m>0,\mathrm{\forall }n\ge 0,$$

where m is a constant. Then

• (a) the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to the common fixed point q of T and S. Also,

  $$\parallel x_{n+1}-q\parallel \le (1-sm)\parallel x_n-q\parallel ,\mathrm{\forall }n\ge 0,$$

  (b)

  $$\parallel y_{n+1}-q\parallel \le (1-sm)\parallel y_n-q\parallel +{\epsilon }_n,\mathrm{\forall }n\ge 0,$$

• (c) ${lim}_{n\to \mathrm{\infty }}{y}_n=q$ implies that ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$.

Example 14 Let $\mathbb{R}$ denote the set of real numbers with the usual norm, $K=\mathbb{R}$, and define $T,S:\mathbb{R}\to \mathbb{R}$ by

$$Tx=\frac{2}{5}{sin}^{2}x,\text{and}Sx=\frac{4}{5}x.$$

Set $L=\frac{4}{5}$, $t=\frac{5}{4}$, $s=\frac{1}{400}$. Clearly, $F(T)\cap F(S)=\{0\}$ and

$$|Tx-Ty|\le \frac{2}{5}|sinx-siny||sinx+siny|\le L|x-y|,\mathrm{\forall }x,y\in \mathbb{R}.$$

Clearly both T and S are Lipschitz operators on $\mathbb{R}$.

Also, it follows from (1.1) that

$$\begin{array}{rcl}|(1+r)(x-y)-rt(Tx-Ty)|& \ge & (1+r)|x-y|-rt|Tx-Ty|\\ =& |x-y|+r(|x-y|-t|Tx-Ty|)\\ \ge & |x-y|\end{array}$$

for any $x,y\in \mathbb{R}$ and $r>0$. Thus T is strongly pseudocontractive and Lemma 7 ensures that T is strictly hemicontractive. Put

then it can be easily seen that

$$L[{(1+L)}^2{b}_n^{\mathrm{\prime }}+c_n^{\mathrm{\prime }}+(1+L)(b_n+c_n)]+\frac{c_n^{\mathrm{\prime }}}{b_n^{\mathrm{\prime }}}\le 0.456\le 0.049375,\mathrm{\forall }n\ge 0.$$

It follows from Theorem 9 that the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ defined by (2.1) converges strongly to the common fixed point 0 of T and S in K and the iterative scheme defined by (2.1) is T-stable.